Physics 9D QUIZ 6 NAME
In a stationary state, the spatial part of the wavefunction of a particle is given by
mm) = exp(:c/a), a: 2 o,
and
1,!)(a:)=0, $<0.
Calculate the uncertainty in the particles position.
(You can use the result I? mm exp (bw) dcc =
Physics 9D Mid Term 1
Answer all 5 problems. All 5 carry equal weight. No credit given for answers without work shown.
Some constants: c = 3.0 108 m/s, mass of proton mp = 1.67 1027 Kg, mass of electron me = 9.1 1031
Kg, magnitude of electron charge q = 1
Physics 9D Mid Term 1
Answer all 5 problems. All 5 carry equal weight. No credit given for answers without work shown.
Some constants: c = 3.0 108 m/s, mass of proton mp = 1.67 1027 Kg, mass of electron me = 9.1 1031
Kg, magnitude of electron charge q = 1
PHYSICS 9D FINALS
Please read the questions carefully. All seven questions carry equal points. NO CREDIT WILL BE GIVEN
FOR ANSWERS WITHOUT WORK SHOWN
Please answer every question 1 through 7 on a separate page in your Blue Book.
For each of the following
Ann measures the length of a rod at rest in her frame
y
x
(Ann)
z
y
v
(Bob)
z
x
Ann measures the length of a rod at rest in her frame
event #1 occurs at t1
y
x
(Ann)
event #1: flash occurs when Bobs
sparker crosses one end of the rod
z
y
v
(Bob)
z
x
Ann m
force diagram:
N
contact force on you causes
centripetal acceleration:
F
N mg
2 r = ac = net = =
=g
m m
m
r
So in an effort to trick you into
thinking you were on Earth, the aliens
cleverly adjusted the rotational speed
of the space station so that combi
simultaneous events for Ann
synchronizing flash
flash occurs at t = 0
y
x
(Ann)
x = L
x = +L
z
time of event #1 = time of event # 2 =
L
c
y
(Ann)
x
z
event #1: detector #1 receives flash
event #2: detector #2 receives flash
A Classical Physics Review
for Modern Physics
This material is written for the student taking modern physics. It is intended as a review of general principles
of classical physics, concentrating on topics most important to modern physics, some of which ma
hf
photoelectric effect: energy accounting
KE
energy in = energy out
incoming energy of photon = work function of metal + max kinetic energy of electron
hf = +KEmax
part 1: both barn doors open
according to farmer
according to farmers wife
event #1: flash occurs when leading
end of the ladder exits the barn
event #2: flash occurs when trailing
end of the ladder enters the barn
Part of the resolution to this paradox i
Physics 9D Homework Assignment #1
CE 1
In our discussion of the twin paradox, we decided that the reason the twins age different amounts between the events of
departure and arrival was that there is an asymmetry the twin that left on the journey accelerat
Bobs view of the elastic collision (after)
y
y
v
y
v
(Ann)
z
v
x
(Ann)
x
(Ann)
z
x
z
y component of velocity of Bob' s ball = +u
y
y component of velocity of Ann' s ball = u ' y
(Bob)
x
z
We know from the symmetry of the situation that both Ann and Bob th
now for a microscopic view
v
v
+
+
+
r
!
+ v=0
Q
+ +
+
charges are equally-spaced
FE=0 (electric field is zero)
FB=0 (v=0 in magnetic field)
y
(Ann)
z
+
+ +
r
+
Q
y
x
(Bob)
+
+ +
v
() charges far apart
(+) charges close together
FE0 (electric field
event spacetime coordinates : x, y, z, t
event spacetime coordinates : x ', y', z ', t '
y
y
event
v
(Ann)
x
(Bob)
z
x
z
lorentz transformation
v
t'= v t 2 x
c
x ' = v ( x vt )
y' = y
z' = z
rewritten to clearly show spacetime mixing when going from An
y
y
event
x
!
r
y
x
y
!
y = r sin
r
!#"#$
x = r cos
x
y' =
(
r sin
x ' = r cos ( ) r [ cos cos + sin sin ] = ( r cos ) cos + ( r sin ) sin
x ' = x cos + ysin
y' = r sin ( ) r [ cos sin + sin cos ] = ( r cos ) sin + ( r sin ) cos
y' = x sin + y cos
# s
V=2v
Both clocks measure the proper time between the
events of the thrown clocks departure and return,
since both events occur at the same place in both
frames. As we saw in discussion, the proper time
between two events is maximal when the events
occur i
11. In a conductor, the highest energy electrons have freedom to gain energy because they have many quantum-mechanically allowed
states above them at higher energy. Nevertheless, as temperature increases, irregularities due to thermal motion of the ions
i
1.
Unless the confines of the particle are small, compared to its wavelength, it will not behave as a wave. And unless it is bound, it
will not form standing waves, which are the basis of quantization.
6.
It never is. Were it zero, the probability of find
1.
It is a quantity that takes on different discrete values, often integral, each of which corresponds to a different value of some
physical quantity (which is accordingly quantized). Quite often it arises from imposing physical conditions, such as contin
11.
13.
21.
26.
23.
36.
44.
46.
51.
The reason is that it" the intensity is low enough that multiple “particles’1 are not in the apparatus at once, then the conclusion
that each interferes with itself is greatly strengthened.
The spot establishes its loca
1T.
23.
23.
31.
35.
4I.
IQuite a hit. No matter how few ultraviolet photons there are. emh has quite a bit more energy than a Sill] nm photon. so it will
produce an electron able to surmount the electrostatic barrier by a considerable amountﬂbe stopping p
11.
1:5.
515.
6-5.
T2.
rs.
Ti.
No. for if we consider a frame in which the initial object is at rest, kinetic energy clearly increases, so rnass would have to
decrease.
No electron travels from one side of the screen to the other. Nothing that can have an
4.
1?.
22.
24.
33.
(a351omr.[h} Later. {e} Ground observers must see my eloelt running slowly, so their eloelt at ‘1’ must be ahead. I don't see
clocks}: and Y as synchronized, so when 1 pass by x,the clock at Y certainly does not read eem,soeven though i
Chu watches elastic collision (after)
y
y
(Ann)
x
z
(Ann)
x
z
y
(Ann)
x
(Bob)
x
z
y
(Bob)
z
y
x
y
(Bob)
x
z
z
y
(Chu)
x
z
Clearly Chu sees a completely symmetric collision: the balls have equal masses, have equal
x-components of velocity, and have equal y
selection rules: allowed and forbidden transitions between 3 lowest energy states of hydrogen
3s n = 3, l = 0, m = 0
n = 3, l = 1, m = +1
3p n = 3, l = 1, m = 0
n = 3, l = 1, m = 1
n = 2, l = 1, m = +1
2 p n = 2, l = 1, m = 0
n = 2, l = 1, m = 1
2s
1s
quantized angular momentum for l = 1 quantum number
!"
"#
l=1
2
2
L = L = l ( l + 1) ! L = 2!
z
+!
Lz = m! = 0
!
y
x
uncertainty in angular momentum
exhibited in x and y components